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il63 [147K]
3 years ago
15

(Atwood’s Machine): Two masses, 9 kg and 12 kg, are attached by a lightweight cord and suspended over a frictionless pulley. Whe

n released, find the acceleration of the system and the tension in the cord.

Physics
2 answers:
Fittoniya [83]3 years ago
5 0

Answer:

Acceleration = 1.428m/s2

Tension = 102.85N

Explanation:

The detailed solution is attached

USPshnik [31]3 years ago
5 0

Answer:

The acceleration of the system is 1.401 m/s² and

The tension in the cord is 100.902 N

Explanation:

Let the 9 kg mass be m

Let the 12 kg mass be M

By Newton's second law of motion we have

For the 9 kg mass, T - mg = ma and for the 12 kg mass we have T - Mg  = -Ma

Here we took the upward acceleration as positive a of the 9 kg mass and the downward acceleration of the 12 kg mass as -a

Solving for T for the 9 kg mass we have

T = mg + ma

Substituting  the value of T in to the 12 kg mass equation, we have

mg + ma - Mg = -Ma or  a = (\frac{M-m}{M+m} )g therefore the acceleration is

1.401 m/s²

and the tension is T = mg + ma = 9×(9.81+1.401) = 100.902 N

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You are conducting an experiment inside a train car that may move along level rail tracks. A load is hung from the ceiling on a
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b

c

e

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Explanation:

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A proton (mass=1.67x10^-27 kg, charge= 1.60x10^-19 C) moves from point A to point under the influence of an electrostatic force
Tom [10]

Answer:

VB -  VA  =  - 33.4

Explanation:

Generally the workdone in moving the proton is mathematically represented as

     W  =  KE_f  - KE_i

Where KE_i \ and \  KE_f \  are\  the\  initial  \  and  \  final \  kinetic \  energy

So

    KE_i  =  \frac{1}{2} m v_a^2

Here v_a is the velocity at A with value  50 m/s

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    KE_i  =  \frac{1}{2} (1.67*10^{-27}) * 50^2

    KE_i  = 2.09 *10^{-24} \  J

Also  

     KE_f  =  \frac{1}{2} m v_b^2

Here v_a is the velocity at A with value 80 km/s = 80000 m/s

=>   KE_f  =  \frac{1}{2} (1.67*10^{-27}) * 80000^2

=>   KE_f  = 5.34 *10^{-18} \  J

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    W  =   5.34 *10^{-18}  - 2.09 *10^{-24}

     W  =   5.34 *10^{-18}  m/s

Now this workdone is also mathematically represented as

     W =  q *  V

So  

    q *  V =   5.34 *10^{-18}

Here  q =  1.60*10^{-19} C

So

        V =   \frac{5.34 *10^{-18} }{1.60*10^{-19}}

         V =   33.4 \  V

Generally proton movement is in the direction of the electric field it means that  VA>VB

So

    VB -  VA  =  - 33.4

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